quantum field theory - Feynman $ivarepsilon$-prescription in path integral by adding an imaginary part to time

It is known that the well-definiteness of the path integral leads to the Feynman's $i\varepsilon$-prescription for the field propagator. I've found many ways of showing this in the literature, but it is precisely the way that I have learned in my QFT course (and which I have not found in literature) that I do not understand.

Context of the problem

Considering the case of a real scalar field for simplicity, one has that the following path integral (evaluated at asymptotic times)

$$\begin{equation} \lim_{T \rightarrow \infty}\int_{\phi(-T, \vec{x})}^{\phi(T,\vec{x})} \mathcal{D}\phi \ \text{exp} \left( i \int^T_T dt \int d³ x \ ( \mathcal{L}+J \phi) \right)\tag{1} \end{equation}$$

can be expressed as

\begin{equation} \lim_{T \rightarrow \infty} \sum_{m, n} e^{-i\left(E_n+E_m \right)T} <\phi, T|n, T>_J _J \tag{2} \end{equation}

where $|n>_J$ are eigenstates of the hamiltonian $H$ in the pressence of the source $J$. In order to make this oscillatory exponential converge (and properly define the path integral) one adds to $T$ a small imaginary part $T \rightarrow T(1-i\varepsilon)$. With this, one writes the vacuum persistence amplitude as

$$\begin{equation} <0|0>_J = \frac{1}{N} \lim_{\varepsilon \rightarrow 0} \ \lim_{T \rightarrow \infty(1-i\varepsilon)} \int \mathcal{D} \phi \ \text{exp} \left( i \int_{-T}^T dt \int d³x (\mathcal{L}+J\phi)\ \right) \equiv \frac{1}{N} Z[J]\tag{3} \end{equation}$$ where the constant $N$ is typically taken to be $N=Z[0]$.

My problem

In order to relate $Z[J]$ with the Feynman propagator $D_{F}(x-y)$, one typically writes the argument of the exponential in the Fourier space, then makes the change of variable

$$\hat{\phi}(p)'=\hat{\phi}(p)+(p²-m²)^{-1} \hat{J}(p)\tag{4}$$ (which leaves $\mathcal{D}\phi'=\mathcal{D}\phi$) to get

$$\begin{equation} Z[J]=Z[0] \text{exp} \left(-\frac{i}{2} \int \frac{d⁴ p}{(2\pi)⁴} \hat{J}(p) \frac{1}{p²-m²} \hat{J}(-p)\right).\tag{5} \end{equation}$$ Here I'm using the $(+,-,-,-)$ Minkowski sign convention.

Now here I've been told that the fact of replacing $T\rightarrow (1-i\varepsilon)T$ to define the path integral is equivalent in Fourier space as $p⁰ \rightarrow (1+i\varepsilon)p⁰$. With this, one gets the correct $i \varepsilon$ Feynman prescription for the propagator $D_F(x-y)$

$$\begin{equation} Z[J]=Z[0] \text{exp} \left(-\frac{i}{2} \int \frac{d⁴ p}{(2\pi)⁴} \hat{J}(p) \frac{1}{p²-m²+i\varepsilon} \hat{J}(-p) \right) = Z[0] \text{exp} \left(-\frac{1}{2} \int d⁴y \int d⁴x \ J(x) D_F(x-y) J(y) \right). \tag{6} \end{equation}$$

And this is the part that I don't get at all, I've tried but I don't see how the fact that $T\rightarrow (1-i\varepsilon)T$ leads in the previous approach to $p⁰ \rightarrow (1+i\varepsilon)p⁰$ and therefore to the Feynman prescription. I'm having nightmares with this, any help would be really appreciated.

NOTE: I use the following convention for the Fourier transform

$$\begin{equation} \hat{\phi}(p)=\int d⁴ x \ \phi(x) e^{-ip \cdot x}\tag{7} \end{equation}$$

so that

$$\begin{equation} \phi(x)= \int \frac{d⁴p}{(2 \pi)⁴} \ \hat{\phi}(p) e^{+ip \cdot x}.\tag{8} \end{equation}$$